Let f be continuous on [a,b] and differential on (a,b). Prove that if a >= 0 there are x1, x2, x3 $\in$ (a,b) such that
$$f'(x_1) = (b+a){f'(x_2)\over 2x_2} = (b^2 + ba + a^2){f'(x_3)\over3x_3^2}$$
I think this problem will use the generalized mean value theorem to solve. However, I don't know how to apply it. Can you suggest a solution? Thank you very much!
By the mean value theorem,
$$\exists x_1 \in (a,b)\,: \quad \frac{f(b)-f(a)}{b-a} =f'(x_1) \,. \tag{1} $$ Consider $g:[a^2,b^2] \to {\mathbb R}$ defined by $g(x)=f(\sqrt{x})$. Applying the mean value theorem to $g$, we infer that there is some $t \in (a^2,b^2)$ such that $$\frac{g(b^2)-g(a^2)}{b^2-a^2} =g'(t) \,,$$ or equivalently, writing $ x_2:=\sqrt{t}$, $$\frac{f(b)-f(a)}{(b-a)(b+a)}= \frac{f'(\sqrt{t})}{2\sqrt{t}}=\frac{f'(x_2)}{2x_2} \,. \tag{2} $$
Similarly, apply the mean value theorem to $h:[a^3,b^3] \to {\mathbb R}$ defined by $h(x)=f(x^{1/3})$. We deduce that there is some $s \in (a^3,b^3)$ such that $$\frac{h(b^3)-h(a^3)}{b^3-a^3} =h'(s) \,,$$ or equivalently, writing $ x_3:= {s^{1/3}}$, $$\frac{f(b)-f(a)}{(b-a)(b^2+ba+a^2)}= \frac{f'(s^{1/3})}{3s^{2/3}} =\frac{f'(x_3)}{3x_3^2} \,. \tag{3} $$